Problem: A group of 25 friends were discussing a large positive integer. ``It can be divided by 1,'' said the first friend. ``It can be divided by 2,'' said the second friend. ``And by 3,'' said the third friend. ``And by 4,'' added the fourth friend. This continued until everyone had made such a comment. If exactly two friends were incorrect, and those two friends said consecutive numbers, what was the least possible integer they were discussing?
Explanation: Let $N$ denote the large positive integer that everyone is discussing.

The two incorrect numbers are consecutive numbers.  To get the smallest possible value of $N$, we must maximize the incorrect numbers. As such, we should start with the highest possible incorrect numbers and work down.

Suppose the two incorrect numbers are 24 and 25.  Then $N$ must still be divisible by $1, 2, 3, \dots, 23.$  This means $N$ is divisible by 3 and 8, so $N$ is divisible by $3 \cdot 8 = 24$, contradiction.  So the two incorrect numbers cannot be 24 and 25.  We can eliminate the other high cases similarly.

One of the incorrect numbers cannot be 22, because $N$ would still be divisible by 2 and 11.

One of the incorrect numbers cannot be 20, because $N$ would still be divisible by 4 and 5.

One of the incorrect numbers cannot be 18, because $N$ would still be divisible by 2 and 9.

On the other hand, suppose the incorrect numbers were 16 and 17.  Then $N$ would still be divisible by $1, 2, 3, \dots, 15, 18, 19, \dots, 25$.  The lcm of these remaining numbers is
\[2^3 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 19 \cdot 23 = 787386600,\]which is not divisible by 16 or 17.  Thus, the incorrect numbers can be 16 and 17, and the smallest possible value of $N$ is $\boxed{787386600}$.